Question

If $\sin x-\sin y=\frac{1}{2}$ and $\cos x-\cos y=11$, then $\tan (x+y)=$ _____

• A. $\frac{3}{8}$
• B. $-\frac{3}{8}$
• C. $\frac{4}{3}$
• D. $-\frac{4}{3}$
• E. $\frac{44}{483}$

Answer 1

Answer: (E)

$\frac{44}{483}$

$\sin x-\sin y=\frac{1}{2}$

$\Rightarrow 2\sin \left(\frac{x-y}{2}\right)\cos \left(\frac{x+y}{2}\right)=\frac{1}{2}...\left(1\right)$

$\cos x-\cos y=11$

$\Rightarrow -2\sin \left(\frac{x-y}{2}\right)\sin \left(\frac{x+y}{2}\right)=11...\left(2\right)$

Dividing equation (2) by (1), we get

$\tan \left(\frac{x+y}{2}\right)=-22$

Now, $\tan \left(2A\right)=\frac{2\tan A}{1-{\tan }^{2}A}$

$\therefore \tan (x+y)=\frac{2\times (-22)}{1-(-22{)}^2}=\frac{-44}{1-484}$

$=\frac{44}{483}$